nLab reflexive coequalizer

Reflexive coequalisers

Reflexive coequalisers



A reflexive pair is a parallel pair f,g:ABf,g\colon A\rightrightarrows B having a common section, i.e. a map s:BAs\colon B\to A such that fs=gs=1 Bf \circ s = g \circ s = 1_B. A reflexive coequalizer is a coequalizer of a reflexive pair. A category has reflexive coequalizers if it has coequalizers of all reflexive pairs.

Dually, a reflexive coequalizer in the opposite category C opC^{op} is called a coreflexive equalizer in CC.


Reflexive coequalizers should not be confused with split coequalizers, a distinct concept.


Any congruence is a reflexive pair, so in particular any quotient of a congruence is a reflexive coequalizer.



If TT is a monad on a cocomplete category CC, then the category C TC^T of Eilenberg Moore algebras is cocomplete if and only if it has reflexive coequalizers. This is the case particularly if TT preserves reflexive coequalizers.

This is due to (Linton).


Suppose C TC^T has reflexive coequalizers. Then C TC^T certainly has coproducts, because if A iA_i is a collection of TT-algebras, then we can form the coequalizer in C TC^T of the reflexive pair

iFUFUA i iFUεA i iεFUA i iFUA i \sum_i F U F U A_i \underoverset {\sum_i F U \varepsilon A_i} {\sum_i \varepsilon F U A_i} {\rightrightarrows} \sum_i F U A_i

using the fact that the displayed coproducts exist because, for example,

iFUA iF( iUA i)\sum_i F U A_i \cong F(\sum_i U A_i)

since the left adjoint FF preserves coproducts, assumed to exist in CC. That this reflexive coequalizer is the coproduct iA i\sum_i A_i in C TC^T is routine.

Finally, a category with coproducts and reflexive coequalizers is cocomplete. It suffices that general coequalizers exist, but it is easily seen that if

f,g:ABf, g \colon A \stackrel{\to}{\to} B

is a parallel pair, then the coequalizer of the reflexive pair

A+B(g,1 B)(f,1 B)BA + B \stackrel{\overset{(f, 1_B)}{\to}}{\underset{(g, 1_B)}{\to}} B

(note both maps are retracts of the inclusion BA+BB \to A + B) also exists, and gives the coequalizer of the first pair. This completes the proof.


If F:C×DEF\colon C\times D\to E is a functor of two variables which preserves reflexive coequalizers in each variable separately (that is, F(c,)F(c,-) and F(,d)F(-,d) preserve reflexive coequalizers for all cCc\in C and dDd\in D), then FF preserves reflexive coequalizers in both variables together.


This is emphatically not the case for arbitrary coequalizers.


of proposition Suppose given two reflexive coequalizers

c 0c 1c 2c_0 \stackrel{\to}{\to} c_1 \to c_2
d 0d 1d 2d_0 \stackrel{\to}{\to} d_1 \to d_2

and let c ijc_{i j} denote F(c i,d j)F(c_i, d_j) so that we have a diagram

c 00 c 01 c 02 c 10 c 11 c 12 c 20 c 21 c 22\array{ c_{0 0} & \stackrel{\to}{\to} & c_{0 1} & \to & c_{0 2} \\ \downarrow \downarrow & & \downarrow \downarrow & & \downarrow \downarrow \\ c_{1 0} & \stackrel{\to}{\to} & c_{1 1} & \to & c_{1 2} \\ \downarrow & & \downarrow & & \downarrow \\ c_{2 0} & \stackrel{\to}{\to} & c_{2 1} & \to & c_{2 2} }

in which all rows and columns are reflexive coequalizers (using preservation of reflexive coequalizers in separate variables), and all squares are serially commutative. According to Toposes, Triples, Theories, lemma 4.2 page 248, the diagonal is also a (reflexive) coequalizer, as claimed. (See also the lemma on page 1 of Johnstone’s Topos Theory.)

Proposition is particularly interesting when FF is the tensor product of a cocomplete closed monoidal category CC. In this case it implies that the free monoid monad on such a category preserves reflexive coequalizers, and thus (by Linton’s theorem) the category of monoid objects in CC is cocomplete.


Reflexive coequalizers in Set commute with finite products:

the nn-fold product functors Set nSetSet^n \stackrel{\prod}{\to} Set preserve reflexive coequalizers.


This follows from prop. as well as from the fact that the diagram category {0d 1s 0d 01}\{ 0 \stackrel{\overset{d_0}{\to}}{\stackrel{\overset{s_0}{\leftarrow}}{\underset{d_1}{\to}}} 1\} with d 0s 0=d 1s 0=idd_0 \circ s_0 = d_1 \circ s_0 = id is a sifted category.

Of course, the diagonal functor Δ:SetSet n\Delta: Set \to Set^n, being left adjoint to the product functor, preserves reflexive coequalizers; therefore the composite

SetΔSet:xhom(n,x)Set \stackrel{\prod \Delta}{\to} Set: x \mapsto \hom(n, x)

also preserves reflexive coequalizers.

This has a further consequence which is technically very convenient:


If TT is a finitary monad on SetSet, then TT preserves reflexive coequalizers.


We have a coend formula for TT:

T() nFinT(n)×hom(n,)T(-) \cong \int^{n \in Fin} T(n) \times \hom(n, -)

and since this is a colimit of functors hom(n,)\hom(n, -) which preserve reflexive coequalizers, TT must also preserve reflexive coequalizers.

Since finitary monads TT preserve reflexive coequalizers, it follows that the monadic functor U:Set TSetU \colon Set^T \to Set reflects reflexive coequalizers, and so since SetSet has reflexive coequalizers, Set TSet^T must as well. Therefore, by proposition , Set TSet^T is cocomplete. This is actually true for infinitary monads TT on SetSet as well, at least if we assume the axiom of choice (see here for a proof), but the argument just given is a choice-free proof for the case of finitary monads.


See also


Last revised on August 22, 2023 at 15:12:34. See the history of this page for a list of all contributions to it.